What is the integral of x(1-x)^{99} ?

The integral of x(1-x)^{99} is 1/101 (1-x)^{101}-1/100 (1-x)^{100}+C

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integral of x(1-x)^{99}

\int x (1 - x)^{99} d x

\frac{1}{101} (1 - x)^{101} - \frac{1}{100} (1 - x)^{100} + C

Solution steps

\int x (1 - x)^{99} d x

Apply u-substitution

: \int - (- u + 1) u^{99} d u

= \int - (- u + 1) u^{99} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int (- u + 1) u^{99} d u

Expand

(- u + 1) u^{99} : - u^{100} + u^{99}

= - \int - u^{100} + u^{99} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{100} d u + \int u^{99} d u)

\int u^{100} d u = \frac{u ^{101}}{101}

\int u^{99} d u = \frac{u ^{100}}{100}

= - (- \frac{u ^{101}}{101} + \frac{u ^{100}}{100})

Substitute back

u = 1 - x

= - (- \frac{( 1 - x ) ^{101}}{101} + \frac{( 1 - x ) ^{100}}{100})

Simplify

- (- \frac{( 1 - x ) ^{101}}{101} + \frac{( 1 - x ) ^{100}}{100}) : \frac{1}{101} (1 - x)^{101} - \frac{1}{100} (1 - x)^{100}

= \frac{1}{101} (1 - x)^{101} - \frac{1}{100} (1 - x)^{100}

Add a constant to the solution

= \frac{1}{101} (1 - x)^{101} - \frac{1}{100} (1 - x)^{100} + C

\int \frac{x}{x ^{4} + 4 x ^{2} + 5} d x

x \to 2 + lim (\frac{2 - x}{∣ x - 2 ∣})

t \to 0 lim (\frac{e ^{\frac{1}{t}}}{t})

\frac{d y}{d x}, 2 x^{3} - 5 x y - y^{2} = 3

x \to - 2 lim ((6 x + 1) (2 x - 3))

What is the integral of x(1-x)^{99} ?
The integral of x(1-x)^{99} is 1/101 (1-x)^{101}-1/100 (1-x)^{100}+C